Let m be the smallest positive integer such that the coefficient of x 2 in the expansion of 1 + x 2 + 1 + x 3 + … + 1 + x 49 + 1 + m x 50 is 3 n + 1 51 C 3 for some positive integer n. Then the value of n is

Step-by-step solution

Coefficient of x 2 in the expansion of 1 + x 2 + 1 + x 3 + … 1 + x 49 + 1 + m x 50 is 2 C 2 + 3 C 2 + … 49 C 2 + 50 C 2 m 2 = 3 n + 1 51 C 3 ⇒ 3 C 3 + 3 C 2 + … 49 C 2 + 50 C 2 m 2 = 3 n + 1 51 C 3 (Use n C r + n C r + 1 = n + 1 C r + 1 ) ⇒ 50 C 3 + 50 C 2 m 2 = 3 n + 1 51 C 3 50.49.48 6 + 50.49 2 m 2 = 3 n + 1 51.50.49 6 m 2 = 51 n + 1 must be a perfect square By trial ⇒ n = 5 and m = 16 ( M , n ∈ N ) ⇒ n = 5